Direction: Study the given information carefully and answer the questions that follow:

A basket contains 5 red, 4 blue, 3 green stones.

If four balls are picked at random, what is the probability that two are blue and two are red?

Probability if two are Blue and two are Red

[(^{4}C_{2}*^{5}C_{2})/^{12}C_{4}] = (6*10)/495 → 4/33

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women

In how many ways this can be done if at least 5 men to have to be included in a committee

Number of Ways when if at least 5 men include in committee = ^{7}C_{5}*^{6}C_{5}+^{7}C_{6}*^{6}C_{4}+^{7}C_{7}*^{6}C_{3}

21*6+7*15+1*20 = 251

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Directions: In the following questions two equations numbered I and II are given. You have to solve both the equations.

I. 5x2- 24x +16 = 0

II. 5y2 + 29y + 20 = 0

**A. ** X>Y

**B. ** X ≥Y

**C. ** X
**D. ** X≤Y

**E. ** X = Y or the relationship cannot be established

**Answer : ****Option A**

**Explaination / Solution: **

I. 5x2- 24x +16 = 0

II. 5y2 + 29y + 20 = 0

I. 5x^{2}- 24x +16 = 0

x = 4/5, 4

II. 5y^{2} + 29y + 20 = 0

y = (-4/5, -5)

So X>Y

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Direction: Study the following information carefully and answer the questions that follow:

A committee of 10 persons is to be formed from 7 men and 6 women.

In how many ways of these committee the women are in majority

Number of ways when women are majority in committee =

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Direction: Study the following information carefully and answer the questions that follow:

A committee of five members is to be formed out of 3 trainees, 4 professors and 6 engineers.

In how many ways this can be done if at least 2 trainees to have to be included in a committee

If at least 2 trainees include in committee

= 405

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A coin is tossed three times, E : at most two tails , F : at least one tail. Find P(E|F)

**A. ** 4/7

**B. ** 3/7

**C. ** 6/7

**D. ** 2/7

**Answer : ****Option C**

**Explaination / Solution: **

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If P(A) =, P(B) = and P(A ∪ B) = find P(A∩B)

**A. ** 4/11

**B. ** 4/13

**C. ** 5/11

**D. ** 5/17

**Answer : ****Option A**

**Explaination / Solution: **

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Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, find P (E|F) and P(F|E)

**A. ** P (E|F) =,P(F|E) =
**B. ** P (E|F) =,P(F|E) =
**C. ** P (E|F) =,P(F|E) =
**D. ** P (E|F) =,P(F|E) =
**Answer : ****Option A**

**Explaination / Solution: **

We have ,

P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then ,

We have ,

P(E) = 0.6, P(F) = 0.3 and P(E ∩ F) = 0.2, then ,

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A dice has 3 faces each bearing ‘ 2 ‘ and three faces each bearing ‘ 6 ‘. It is rolled once. The probability of showing up ‘a six ‘ is

**A. ** 1

**B. ** none of these

**C. ** 1/6

**D. ** 1/2

**Answer : ****Option D**

**Explaination / Solution: **

Total no. of outcomes = 6={2,2,2,6,6,6}

p(getting a six)=3/6=1/2

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The probability that a card drawn at random from a pack of 52 cards is a king or a heart is

**A. ** 1/ 52

**B. ** 1/ 13

**C. ** 16/ 52

**D. ** 1/ 4

**Answer : ****Option C**

**Explaination / Solution: **

No Explaination.

No Explaination.

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